$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$

Let $a, b, c$ be positive real numbers such that $a\geq b\geq c$ and $abc=1$ prove that $$\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\geq \frac{3}{\sqrt{2}}$$

Solution 1:

Using Hölder's inequality we have:

$$\left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^{2/3} (a(a+b)+b(b+c)+c(c+a))^{1/3}\geq a+b+c.$$

i.e.

$$\left(\frac{a}{\sqrt{a+b}}+\frac{b}{\sqrt{b+c}}+\frac{c}{\sqrt{c+a}}\right)^{2} \geq \frac{(a+b+c)^{3}}{a^2+b^2+c^2+ab+bc+ca}.$$

We have to prove that:

$$\frac{(a+b+c)^{3}}{a^2+b^2+c^2+ab+bc+ca} \geq \frac{9}{2}.$$

i.e. $$2(a+b+c)^3\geq9\left(a^2+b^2+c^2+ab+bc+ca\right). \tag{1}$$

Let $p=a+b+c$ and $q=ab+bc+ca$ and using that $abc=1$ and $AM-GM$ we obtain that $q \geq 3$.

Inequality $(1)$ is equivalent to:

$$2p^3 \geq 9\left(p^2-2q+q\right) \Leftrightarrow 2p^3+9q \geq 9p^2.$$

Applying $AM-GM$ we obtain

$$2p^3+9q \geq 2p^3+27=p^3+p^3+27 \geq 3\cdot \sqrt[3]{27p^6}=9p^2,$$ as required.

Solution 2:

Let $a=b$. Then $c=\frac{1}{a^2}$ and the formula is : $$f(a)=\sqrt{\frac{a}{2}}+\sqrt{a+\frac{1}{a^2}}$$ $$f'(a)=\frac{1}{2\sqrt{2a}}+(\frac{1}{2}-\frac{1}{a^3}).\frac{1}{\sqrt{a+\frac{1}{a^2}}}$$

The only root in $[0,\infty)$ of f'(a) is 1, hence $f(1)=\frac{3}{\sqrt{2}}$ is a minimum.

Now, what happens if $a\neq b$ ? Use the same method, but say $b=k.a+(1-k)$ (so let $k=\frac{b-1}{a-1}$ be a constant, and if $a=1$ exchange $a$ and $c$). Then $c=\frac{1}{ka^2+a(1-k)}$ and :

$$f_k(a)=\frac{a}{\sqrt{a.(1+k)+(1-k)}}+\frac{ka+(1-k)}{\sqrt{ka+(1-k)+\frac{1}{ka^2+a(1-k)}}}+\frac{\frac{1}{ka^2+a(1-k)}}{\sqrt{a+\frac{1}{ka^2+a(1-k)}}}$$

Once again, obtain $f'_k$ and show that its only root is $1$ (This is quite technical, use mathematica ?). I agree there should have something more simple.